The number
\[e^{7\pi i/60} + e^{17\pi i/60} + e^{27 \pi i/60} + e^{37\pi i /60} + e^{47 \pi i /60}\]is expressed in the form $r e^{i \theta}$, where $0 \le \theta < 2\pi$.  Find $\theta$.
Answer: Let's locate these numbers in the complex plane before adding them. Since $e^{i \theta}$ is the terminal point for angle $\theta$ on the unit circle, here are the numbers:
[asy]
size(200); 
import TrigMacros;
rr_cartesian_axes(-2,2,-1,3,complexplane=true, usegrid = false);
pair O = (0,0); 
pair[] Z; 
for (int i = 0; i < 5; ++i)
{
     Z[i] = dir(30i)*dir(12); 
     draw(O--Z[i]); 
     dot(Z[i]); 
} 
label("$e^{7\pi i/60}$", Z[0], dir(Z[0])); 
label("$e^{17\pi i/60}$", Z[1], dir(Z[1])); 
label("$e^{27\pi i/60}$", Z[2], dir(Z[2])); 
label("$e^{37\pi i/60}$", Z[3], NNW); 
label("$e^{47\pi i/60}$", Z[4], NW); 
[/asy] We need to add all $5$ numbers. However, we don't actually need to find the exponential form of the answer: we just need to know argument of our sum, that is, the angle that our sum makes with the positive $x$-axis.

The symmetry of the above picture suggest that we consider what happens if we add up pairs of numbers. For example, let's try adding $e^{7\pi i/60}$ and $e^{47\pi i /60}$ head to tail:
[asy]
size(200); 
import TrigMacros;
rr_cartesian_axes(-2,2,-1,3,complexplane=true, usegrid = false);
pair O = (0,0); 
pair[] Z; 
for (int i = 0; i < 5; ++i)
{
     Z[i] = dir(30i)*dir(12); 

} 
draw(O--Z[0], blue);
draw(O--Z[4]);
draw(Z[4]--Z[0]+Z[4], blue); 
draw(O--Z[0]+Z[4]); 
dot("$e^{7\pi i/60}$", Z[0], dir(Z[0])); 
dot("$e^{47\pi i/60}$", Z[4], NW); 
dot("$e^{7\pi i/60} + e^{47\pi i/60}$", Z[4]+Z[0], N); 
[/asy]
Since $|e^{7\pi i/60}| = |e^{47\pi i/60}| = 1$, the parallelogram with vertices at $0, e^{7\pi i/60}, e^{47 \pi i/60}$ and $e^{7\pi i/ 60} + e^{47 \pi i/60}$ is a rhombus. That means that the line segment from $0$ to $e^{7\pi i/ 60} + e^{47 \pi i/60}$  splits the angle at $0$ in half, which means that the argument of $e^{7\pi i/60} + e^{47 \pi i/60}$ is the average of the arguments of the numbers being added, or in other words is
\[\dfrac{1}{2} \left( \dfrac{7\pi}{60} + \dfrac{47\pi}{60}\right) = \dfrac{27 \pi}{60} = \dfrac{9\pi}{20}.\]That means that
\[ e^{7\pi i/ 60} + e^{47 \pi i/60} = r_1 e^{9 \pi i/20},\]for some nonnegative $r_1$.

Similarly, we can consider the sum $e^{17\pi i/60} + e^{37\pi i/60}$. Here it is in the picture:

[asy]
size(200); 
import TrigMacros;
rr_cartesian_axes(-2,2,-1,3,complexplane=true, usegrid = false);
pair O = (0,0); 
pair[] Z; 
for (int i = 0; i < 5; ++i)
{
     Z[i] = dir(30i)*dir(12); 

} 
draw(O--Z[1], blue);
draw(O--Z[3]);
draw(Z[3]--Z[1]+Z[3], blue); 
draw(O--Z[1]+Z[3]); 
dot("$e^{17\pi i/60}$", Z[1], dir(Z[1])); 
dot("$e^{37\pi i/60}$", Z[3], NW); 
dot("$e^{17\pi i/60} + e^{37\pi i/60}$", Z[3]+Z[1], N); 
[/asy]We again have a rhombus, which again means that the sum of the pair has an argument equal to the average of the arguments. That means that the argument of $e^{17\pi i/60} + e^{37 \pi i/60}$ is the average of the arguments of the numbers being added, or in other words is
\[\dfrac{1}{2} \left( \dfrac{17\pi}{60} + \dfrac{37\pi}{60}\right) = \dfrac{27 \pi}{60} = \dfrac{9\pi}{20}.\]Therefore,
\[ e^{17\pi i/ 60} + e^{37 \pi i/60} = r_2 e^{9 \pi i/20},\]for some nonnegative $r_2$.

Finally, our middle number is $e^{27\pi i/60} = e^{9\pi i/20}$, simplifying the fraction. Now we're adding up three numbers with argument $e^{9\pi i/20}$, which gives another number with the same argument. To be more precise, we have that
\begin{align*} 
e^{7\pi i/60} + e^{17\pi i/60} + e^{27 \pi i/60} + e^{37\pi i /60} + e^{47 \pi i /60} &= (e^{7\pi i/60} + e^{47\pi i/60}) + e^{27 \pi i/60} + (e^{37\pi i /60} + e^{47 \pi i /60})  \\
&= r_1 e^{9\pi i/20} + e^{9\pi i/20} + r_2 e^{9\pi i/20} \\
&= (r_1 +r_2 + 1) e^{9\pi i/20},
\end{align*}which gives that the argument of our sum is $\boxed{\dfrac{9\pi}{20}}$.